Multiple Balls: Disadvantage?

[mas0764]

It doesn't. Both systems give the teams exactly the same odds of winning the #1 seed.

It does. You are myopically looking at a closed system with 48 balls where any one of them could be the last ball standing, and then multiplying by three cause some teams have three balls.

The problem is, if a bunch of balls get pulled from bad teams early (and this is likely due to their higher number of balls available), they end up with equal chances as everyone else to end up with picks in the middle of the draft.

Not even considering the first pick overall, once a bad team misses out on a high one, its odds of getting a pick around 8, 9, 10, etc, should be greatly weighted in its favor over one-ball teams, and should serve to prevent them from ending up with a mid-late round pick.

The backwards method levels the playing field, increasing the chances that a bad team gets stuck with a bad pick.

If you don't understand this, you don't understand stats. Sorry to have to call people out on this, but if you disagree, you're wrong.

 

[Slats432]

Confucious say, Baseball wrong, man with four balls cannot walk.

Confucius also say that man who stands with hands in pockets feel cocky all day.

Same could be said for team with three balls.

You guys and all this ball talk is just nuts.

 

[MaV]

It does. You are myopically looking at a closed system with 48 balls where any one of them could be the last ball standing, and then multiplying by three cause some teams have three balls.

The problem is, if a bunch of balls get pulled from bad teams early (and this is likely due to their higher number of balls available), they end up with equal chances as everyone else to end up with picks in the middle of the draft.

Ah, so you were talking about the whole order. I should have taken a closer look, now I see that on your reply to 'bcrt2000'.

 

[PecaFan]

The problem is, if a bunch of balls get pulled from bad teams early (and this is likely due to their higher number of balls available), they end up with equal chances as everyone else to end up with picks in the middle of the draft.

And the chance of that happening is equal to the opposite situation: a bunch of one ball teams get their balls drawn early in the 1 to 30 normal draft, and the three ball teams end up in the middle of the draft.

It's *possible* that three ball teams get all their balls drawn early. Not likely.

It's just as possible that a three ball team never gets one of it's balls drawn until 30th spot. But it's also not likely.

 

[mas0764]

And the chance of that happening is equal to the opposite situation: a bunch of one ball teams get their balls drawn early in the 1 to 30 normal draft, and the three ball teams end up in the middle of the draft.

The idiotic backwards drafting theory creates situations where a three-ball team is in a 1:1 race down the stretch with one-ball teams. This is not guaranteed, but it is a possibility.

The frontwards drafting theory always gives the edge on a per-pick basis to the bad teams with multiple balls. This is not arguable, because its mathematical fact. If you don't believe me, then feel free to map out every scenario along with percentages in your next post, and in the process prove yourself wrong.

If the idea is to give the advantage to the bad teams, the frontwards drafting theory is the only way that assures this STATISTICAL advantage. End of story.

 

[MaV]

So I don't know statistics that well. Teach me.

What if there was a lottery that had five teams, four with one balls, one with three. Would it be similar to this situation now? Would the results depends on starting from #1 or #5?

 

[bcrt2000]

I find it hard to take your word because no statistician looks at it that way.

The statistical probability of any one ball being pulled out is all that matters and it changes after every pull.


If you start at 30, and for ease's sake, lets say the Rangers and Blackhawks have three balls each, and every other team only has one.... meaning there are 34 balls....

Each team has a 1/34 chance of being picked, except the Rangers and Blackhawks, who have a 3/34 chance of being pulled out.

If the Rangers and Blackhawks aren't picked, they now have a 3/33 chance of being picked out, while the other teams still only have a 1/33 chance of being pulled.

On the next pick, the Rangers first ball comes out.

Now its:

Blackhawks 3/32
Rangers 2/32
Everyone else 1/32

Next its someone else.

Blk 3/31
Rangers 2/31
Everyone 1/31

Eventually, you will reach a point, and given the odds of lasting, probably well before the tenth overall pick, where the Rangers and Blackhawks catch up with all the remaining teams right about 1/16. It is inevitable that they catch up because of their better chance of being pulled out every time.

So instead of the Rangers and Hawks having a 6% chance to get the first pick and everyone else having a 2% chance, they are now on equal footing to get any particular draft pick from here on out, thus offering the Ranger and Hawks no statistical advantage.

The problem with doing it like this is that the Rangers and Hawks now have pretty good odds at being pulled out at 16 or 15.

This isn't fair, because if you do it the other way, it is very difficult for the Rangers or Hawks to not have been pulled by 16, since the odds get BETTER when you work the other way.

Given that we are in the business of helping the bad teams out by weighting the system, the only fair way to do it is to start at the front. It protects the interests of bad teams while maintaining reasonable odds for the good teams.

Hypothesis: forward lottery is equvalent to reverse lottery

Proof:

First, I'm assuming that everyone here understands that if you have a set of n unique balls {1,2,3,4,... n-2, n-1, n}, each ball has an equal probability of being selected at any position (P_i = 1/(n-i+1)), 1<= i <= n, then there are n! possible outcomes, and the probability for each possible outcome is 1/(n!). In addition, the probability of an outcome, is the SAME as the probability of its reverse (ie, if we have 3 balls, and the order the balls are selected is {2,1,3}, the probability of this outcome = 1/3! = 1/(3*2*1) = 1/6. The probability of its reverse, {3,1,2} is also 1/3!, which is 1/6). This is not a proof for this claim, but I'm pretty sure you could look it up in any stats textbook (probably one of the first random variable properties that was ever proved), and while its simple to understand this, its probably a bit harder to prove it, and I'm not going there because this is obvious.

Okay, so we've established the fact that with n unique balls, that its equally likely that any order of n balls being selected is as likely as it's reverse being selected.

Suppose we have 48 balls, each numbered uniquely, ie {1, 2, 3, ... , 47, 48}. Any outcome of the 48 balls is equally likely as it's reverse. So in a forward draft, if we get a certain outcome, the outcome's reverse is as equally likely to occur in a reverse draft.

We can easily assign each team to its proper amount of balls, ie ball #1 = BUF ball #1, ball #2 = BUF ball #2, ball #2 = BUF ball #2, ball #4 = CBJ ball #1, etc

So now we have to show that counting a team's ball first in the forward draft is equivalent to only counting the last ball in the reverse draft. This is pretty obvious-- lets just look at a smaller example, 5 balls with 3 teams (the draft is only for 3 picks)... say the forward order that we picked out the balls is {DET, NYR ball #2, COL, NYR ball #1, NYR ball #3}.. we proved that its probability would be equivalent to a reverse draft order of {NYR #3, NYR #1, COL, NYR #2, DET}. In the forward Draft, the first pick would be Detroit, second pick NYR (NYR #2), third pick COL, then we ignore the last 2 balls. In the reverse Draft, we'd ignore the first NYR ball (NYR #3), ignore the second NYR ball (NYR #1), then COL only has one ball left, so it gets the third pick, then NYR's last ball (NYR #2), so they get second pick, and Detroit has the last ball so they get first pick. You could extend this into some induction proof but I think this is good enough for the purposes of this board to prove that taking a forward order of 48 balls and flipping the order and doing the reverse order elimination draft is the same

Therefore Forward Draft is equivalent to Backwards Draft.


holy crap I can't believe I did that

Theres probably a way to prove it by anaylzing the statistics, but its a long ass way, and frankly, no statistician would do it unless they wanted to give their brain a work out

 

[jekoh]

It does. You are myopically looking at a closed system with 48 balls where any one of them could be the last ball standing, and then multiplying by three cause some teams have three balls.

The problem is, if a bunch of balls get pulled from bad teams early (and this is likely due to their higher number of balls available), they end up with equal chances as everyone else to end up with picks in the middle of the draft.

So what ? That doesn't change the fact that the last ball will win.

if you disagree, you're wrong.

You should be able to tell us what the odds are with the backwards system, then.

For instance, think of a 3-team draw, with two one-ball teams and one two-ball team. What are the odds with both methods, tell me ?

 

[MaV]

You should be able to tell us what the odds are with the backwards system, then.

For instance, think of a 3-team draw, with two one-ball teams and one two-ball team. What are the odds with both methods, tell me ?

I have the odds for a small 5-team system, with one three ball team. There is some striking resemblance on the forward and backwards system odds ; -)

 

[mas0764]

Perhaps I am out of the loop here, (and maybe someone has already addressed this) but if they are going to draw balls backwards (starting with #30 all the way to #1), isn't it going to be a disadvantage to have multiple balls? (assuming once your ball is picked, the rest of your balls are null and void)

Can someone explain?

No; because what they are saying is that if you have multiple balls, you disregard your balls until you draw out your final ball, and that's where you pick, sort of like a last person standing.

 

[ArtG]

No; because what they are saying is that if you have multiple balls, you disregard your balls until you draw out your final ball, and that's where you pick, sort of like a last person standing.

Finally on the bandwagon, eh?

Math is really not a strong point on these boards... so let's just stick to hockey talk

Anyhow, it is in fact true that either way you do the lottery the probabilities remain the same. Forward or reverse should not affect probability unless balls are removed prematurely.

 

[mas0764]

Finally on the bandwagon, eh?

Math is really not a strong point on these boards... so let's just stick to hockey talk

Anyhow, it is in fact true that either way you do the lottery the probabilities remain the same. Forward or reverse should not affect probability unless balls are removed prematurely.

No, I am not on the bandwagon.... still reading that proof.

However, that person was asking if its more likely to hurt a three ball team because one of your three balls get picked out early and that's where you pick.

I was just clarifying that we are saying that you disregard the first balls selected.

 

[Seph]

It does. You are myopically looking at a closed system with 48 balls where any one of them could be the last ball standing, and then multiplying by three cause some teams have three balls.

The problem is, if a bunch of balls get pulled from bad teams early (and this is likely due to their higher number of balls available), they end up with equal chances as everyone else to end up with picks in the middle of the draft.

Not even considering the first pick overall, once a bad team misses out on a high one, its odds of getting a pick around 8, 9, 10, etc, should be greatly weighted in its favor over one-ball teams, and should serve to prevent them from ending up with a mid-late round pick.

The backwards method levels the playing field, increasing the chances that a bad team gets stuck with a bad pick.

If you don't understand this, you don't understand stats. Sorry to have to call people out on this, but if you disagree, you're wrong.

No, actually, you don't understand statistics.

By the time a team is down to 1 ball, there will be a number of teams that have zero balls left also. Thus increasing the odds.

You are looking at this the wrong way. Imagine 48 slots, each with a number on them. Now you have to fill those slots at random with 48 balls. The order you do it in does not matter. If you did it all at once, backwards, forwards, randomly, each of those slots would still have the same 1/48 chance of getting a specific ball. If three of those ball were the identical, each of those slots would have a 3/48 chance. If you go forwards, this is what you're doing, you just ignore the 18 redundant slots.

It seems like the odds change as the draft continues, but that's only the odds for which ball comes next. And yes, the odds would change for that. But if the only thing that matters is the end outcome, the odds remain exactly the same, drawn forward or back.

 

[PecaFan]

The idiotic backwards drafting theory creates situations where a three-ball team is in a 1:1 race down the stretch with one-ball teams. This is not guaranteed, but it is a possibility.

The frontwards drafting theory always gives the edge on a per-pick basis to the bad teams with multiple balls. This is not arguable, because its mathematical fact.

What you continue to miss is that these possibilities exist under the 1 to 30 draft as well. They're just as much mathematical fact as yours.

Just three clicks on my official sim gives this (runs like this are *common*):
http://members.shaw.ca/vorpalbunnies/draftsim/official.html

1. Ott ( 2.1%)
2. Wsh ( 2.1%)
3. Col ( 2.2%)
4. Min ( 4.4%)
5. Ana ( 4.7%)
6. Phx ( 4.9%)
7. Mtl ( 2.6%)
8. SJS ( 2.6%)
9. Nsh ( 5.4%)
10. Det ( 2.9%)
11. Cgy ( 5.9%)
12. Car ( 6.3%)
13. CBJ ( 10.0%)
14. Atl ( 7.4%)
15. LAK ( 8.0%)
16. Pit ( 13.0%)
17. Chi ( 10.0%)
18. Stl ( 5.6%)
19. NYI ( 5.9%)
20. Bos ( 6.3%)
21. Edm ( 13.3%)
22. Buf ( 23.1%)
23. Van ( 10.0%)
24. Tam ( 11.1%)
25. Tor ( 12.5%)
26. NYR ( 42.9%)
27. Fla ( 25.0%)
28. Dal ( 33.3%)
29. Phi ( 50.0%)
30. NJD (100.0%)

Top *three* teams are 1 ballers. Top *12* are one and two ballers, so the highest three baller is 13th. Final three ball teams at 22nd and 26th, after *all ten* two ball teams have had their picks.

Look at the last 13 teams. There's poor Buffalo and NY, 3 ball teams battling it out with one two ball team (Edmonton), and *ten* one ball teams. And six of them manage to *still* get ahead of the Rangers.

How about a different tack, then:

There are 48 horses in a race. You can bet on one horse to win. I can bet on three. You're saying our odds of getting the winning horse are the same? After all, it's a mathmatical fact that two of my horses will not finish first, so we'll be in an equal situation after that?

 

[mas0764]

Say the Rangers have balls 1 and 2. Blackhawks have ball 3. And that's all in the entire draft, no other teams.

We'll call the slots x and y, with x being first chronologically.

If you start from the front with x, the Rangers have a 2/3 chance of winning immediately by having their ball pulled out first in slot x. The Blackhawks have a 1/3 chance. That's all.

But if you start from the back with y, the Rangers only have a 1/3 chance of winning immediately by having the Blackhawks ball pulled out. If the Rangers do not win immediately (ie, ball 1 or 2 is pulled out on the first draw), they then face a 1/2 chance of winning immediately.

They have a 1/3 chance of winning on the first pull, and a 1/2 chance of winning on the second pull, if you work backwards. I forget how you combine those odds.

They have a 2/3 chance of winning on the first pull working forwards.

 

[mas0764]

No, actually, you don't understand statistics.

By the time a team is down to 1 ball, there will be a number of teams that have zero balls left also. Thus increasing the odds.

You are looking at this the wrong way. Imagine 48 slots, each with a number on them. Now you have to fill those slots at random with 48 balls. The order you do it in does not matter. If you did it all at once, backwards, forwards, randomly, each of those slots would still have the same 1/48 chance of getting a specific ball. If three of those ball were the identical, each of those slots would have a 3/48 chance. If you go forwards, this is what you're doing, you just ignore the 18 redundant slots.

It seems like the odds change as the draft continues, but that's only the odds for which ball comes next. And yes, the odds would change for that. But if the only thing that matters is the end outcome, the odds remain exactly the same, drawn forward or back.

Isn't that what matters, what ball comes next?

Cause we're not just trying to figure out who goes first.

 

[PecaFan]

Say the Rangers have balls 1 and 2. Blackhawks have ball 3. And that's all in the entire draft, no other teams.

We'll call the slots x and y, with x being first chronologically.

If you start from the front with x, the Rangers have a 2/3 chance of winning immediately by having their ball pulled out first in slot x. The Blackhawks have a 1/3 chance. That's all.

But if you start from the back with y, the Rangers only have a 1/3 chance of winning immediately by having the Blackhawks ball pulled out. If the Rangers do not win immediately (ie, ball 1 or 2 is pulled out on the first draw), they then face a 1/2 chance of winning immediately.

They have a 1/3 chance of winning on the first pull, and a 1/2 chance of winning on the second pull, if you work backwards. I forget how you combine those odds.

They have a 2/3 chance of winning on the first pull working forwards.

Possible combos of balls drawn in order:
NY1, NY2, Chi1 = Chicago win #1 pick
NY1, Chi1, NY2 = NY win
NY2, NY1, Chi1 = Chicago win
NY2, Chi1, NY1 = NY win
Chi1, NY1, NY2 = NY win
Chi1, NY2, NY1 = NY win

Gee Wally, the Rangers win 4 of the 6 times.

 

[amon]

it doesnt matter who win crosby .The think that i dont like is that the lottery have been made to favorize ny ranger and its not a coincidence

 

[MaV]

They have a 1/3 chance of winning on the first pull, and a 1/2 chance of winning on the second pull, if you work backwards. I forget how you combine those odds.

Doesn't this say enough?

It's 1/3 + (1-(1/3))*1/2. So it's the same.

For a nice five team lottery with one three ball team:

Forward lottery:

Odds for the three ball team getting the pick NR1: 3/7 = 15/35

NR2: 4/6 * 3/6 = 2/7 = 10/35

NR3: 4/7 * 3/6 * 3/5 = 6/35

NR4: 4/7 * 3/6 * 2/5 * 3/4 = 3/35

NR5: 4/7 * 3/6 * 2/5 * 1/4 (* 3/3) = 1/35

Backward lottery:

Odds for the three ball team getting the pick NR5: 3/7 * 2/6 * 1/5 = 1 /35

N4: 4/7 * 3/6 * 2/5 * 1/4 + 3/7 * 4/6 * 2/5 * 1/4 + 3/7 * 2/6 * 4/5 * 1/4 = 3*1/35 = 3/35

(NOTE, we want to calculate just one odd for the right number of one ball teams balls getting picked before the third ball of the three ball team. In this case there is three permutations for those two 3ball balls and the one 1ball ball. In future, just calculating one odd and multiplying it is enough.)

NR3: 6 * (4/7 * 3/6 * 3/5 * 2/4) * 1/3 = 6/35

NR4: 10 * (4/7 * 3/6 * 2/5 * 3/4 * 2/3) * 1/2 = 10/35

NR5: 15* (4/7 * 3/6 * 2/5 * 1/4 * 3/3 * 2/2) * 1/1 = 15/35

So, looks same to me.

 

[Vincent_TheGreat]

holy crap, their drawing the balls 1-30 and announcing on TSN 30-1. Its pretty simple, come on guys.

 

[Briere03]

Alright, here is my theory. Bettman will have a big jug full of balls. The jug will have a horizontal tube attached to the bottom of it, and when Bettman flicks a switch, 30 balls rush out onto the tube. The ball on the farthest left, is Crosby's team. But he will start revealing with the ball farthest right.

This probably makes absolutely no sense, I'm a bit tired.

 

[Patman]

Hypothesis: forward lottery is equvalent to reverse lottery

Proof:

First, I'm assuming that everyone here understands that if you have a set of n unique balls {1,2,3,4,... n-2, n-1, n}, each ball has an equal probability of being selected at any position (P_i = 1/(n-i+1)), 1<= i <= n, then there are n! possible outcomes, and the probability for each possible outcome is 1/(n!). In addition, the probability of an outcome, is the SAME as the probability of its reverse (ie, if we have 3 balls, and the order the balls are selected is {2,1,3}, the probability of this outcome = 1/3! = 1/(3*2*1) = 1/6. The probability of its reverse, {3,1,2} is also 1/3!, which is 1/6). This is not a proof for this claim, but I'm pretty sure you could look it up in any stats textbook (probably one of the first random variable properties that was ever proved), and while its simple to understand this, its probably a bit harder to prove it, and I'm not going there because this is obvious.

Okay, so we've established the fact that with n unique balls, that its equally likely that any order of n balls being selected is as likely as it's reverse being selected.

Suppose we have 48 balls, each numbered uniquely, ie {1, 2, 3, ... , 47, 48}. Any outcome of the 48 balls is equally likely as it's reverse. So in a forward draft, if we get a certain outcome, the outcome's reverse is as equally likely to occur in a reverse draft.

We can easily assign each team to its proper amount of balls, ie ball #1 = BUF ball #1, ball #2 = BUF ball #2, ball #2 = BUF ball #2, ball #4 = CBJ ball #1, etc

So now we have to show that counting a team's ball first in the forward draft is equivalent to only counting the last ball in the reverse draft. This is pretty obvious-- lets just look at a smaller example, 5 balls with 3 teams (the draft is only for 3 picks)... say the forward order that we picked out the balls is {DET, NYR ball #2, COL, NYR ball #1, NYR ball #3}.. we proved that its probability would be equivalent to a reverse draft order of {NYR #3, NYR #1, COL, NYR #2, DET}. In the forward Draft, the first pick would be Detroit, second pick NYR (NYR #2), third pick COL, then we ignore the last 2 balls. In the reverse Draft, we'd ignore the first NYR ball (NYR #3), ignore the second NYR ball (NYR #1), then COL only has one ball left, so it gets the third pick, then NYR's last ball (NYR #2), so they get second pick, and Detroit has the last ball so they get first pick. You could extend this into some induction proof but I think this is good enough for the purposes of this board to prove that taking a forward order of 48 balls and flipping the order and doing the reverse order elimination draft is the same

Therefore Forward Draft is equivalent to Backwards Draft.


holy crap I can't believe I did that

Theres probably a way to prove it by anaylzing the statistics, but its a long ass way, and frankly, no statistician would do it unless they wanted to give their brain a work out

No, that's probably the most straightforward way... and with no statistics. Let a be a set of numeric elements {1,...,n} assigned in a fashion defined by an ommitted algorithm such that each number is assigned to an element in a randomly. Let b a be set of size m<=n which contains a pre-defined arrangement unique numbers {1,...n}. Then the first element that belongs to the set B is the same as the last element of a set defined as a'[n+1-i].

Clearly some a[q] will be the first element in a that is also an element in b. Likewise we know by our ordering principle that this is the same value as a'[n-q]. Since {a[1],...,a[q]}INTERSECT{b[1],...b[m]} has only one element in common (cardinality 1) and a INTERSECT b is B (with cardinarilty m) and {a[1],...,q[q]} and {a[q+1],...,a[n]} are disjoint (and their union whole set of A... these picks are a partitioning of A) then {a[q+1],...,a[n]}INTERSECT{b[1],...b[m]} has cardinality m-1 as these sets must partition b.

returning to our ordering principle {a'[1],...,a'[n-q]}INTERSECT{b[1],...b[m]} has cardinality m-1 (everything else has been chucked) so there is only one element in {a'[n-q+1],...,a'[n]} INTERSECT {b[1],...b[m]}. Clearly this element will be a'[n-q+1] since a'[n-q+1]=a[q]. So the last element selected from b from a reverse ordering is the same as the first element selected from a forward ordering.

It is trivial to extend to all teams since we know that all teams picksets are defined as disjoint sets b1,...,b30. If b1's first element is at b1=q then its last in a reverse ordering is at b1=n-q+1, which is the same since its q from the end. All other picks are chosen in the same fashion so one ordering is simply the mirror of the other.

Lots of nasty set theory... ugh.

 

[bcrt2000]

No, that's probably the most straightforward way... and with no statistics. Let a be a set of numeric elements {1,...,n} assigned in a fashion defined by an ommitted algorithm such that each number is assigned to an element in a randomly. Let b a be set of size m<=n which contains a pre-defined arrangement unique numbers {1,...n}. Then the first element that belongs to the set B is the same as the last element of a set defined as a'[n+1-i].

Clearly some a[q] will be the first element in a that is also an element in b. Likewise we know by our ordering principle that this is the same value as a'[n-q]. Since {a[1],...,a[q]}INTERSECT{b[1],...b[m]} has only one element in common (cardinality 1) and a INTERSECT b is B (with cardinarilty m) and {a[1],...,q[q]} and {a[q+1],...,a[n]} are disjoint (and their union whole set of A... these picks are a partitioning of A) then {a[q+1],...,a[n]}INTERSECT{b[1],...b[m]} has cardinality m-1 as these sets must partition b.

returning to our ordering principle {a'[1],...,a'[n-q]}INTERSECT{b[1],...b[m]} has cardinality m-1 (everything else has been chucked) so there is only one element in {a'[n-q+1],...,a'[n]} INTERSECT {b[1],...b[m]}. Clearly this element will be a'[n-q+1] since a'[n-q+1]=a[q]. So the last element selected from b from a reverse ordering is the same as the first element selected from a forward ordering.

It is trivial to extend to all teams since we know that all teams picksets are defined as disjoint sets b1,...,b30. If b1's first element is at b1=q then its last in a reverse ordering is at b1=n-q+1, which is the same since its q from the end. All other picks are chosen in the same fashion so one ordering is simply the mirror of the other.

Lots of nasty set theory... ugh.

your proof owned my proof

of course yours is more concise & formal, but they say, the shortest proofs, while usually stronger, also usually require much more thought

by the way, i noted one mistake in my proof, i should have said P_i = probability of REMAINING balls being chosen at any position

also, i said that i didn't prove my claim that for any outcome, its equally likely that its reverse will be chosen... well i also showed that each outcome is has a probability of 1/n!, so in essence, the Probability of an outcome = Probability of the reverse, so that was kinda indirectly implied in my proof

 

[ArtG]

your proof owned my proof

of course yours is more concise & formal, but they say, the shortest proofs, while usually stronger, also usually require much more thought

Yes his is definitely a lot more elegant. Reminds of my days battling Discrete Mathematics in first and second year...

 

[bcrt2000]

Yes his is definitely a lot more elegant. Reminds of my days battling Discrete Mathematics in first and second year...

oh the horror of discrete mathematics ... the class average on the midterm in the 2nd year course we had... was like 15% (i'm not joking!)... over 2/3's of the class dropped it including me (good thing it doesn't come on your transcript if you drop before a certain date) ... of course it was a breeze the second time around